How do you divide #(x^2-x-12) -: (x^2-9)/(x^2-3x)#?

1 Answer
EZ as pi
Aug 22, 2017

#=x(x-4)#

Explanation:

Factorise wherever possible.

#(x^2 -x-12) div (x^2-9)/(x^2-3x)#

#=(x-4)(x+3) div((x+3)(x-3))/(x(x-3))#

To divide, multiply by the reciprocal:

#=(x-4)cancel((x+3)) xx (xcancel((x-3)))/(cancel((x+3))cancel((x-3)))#

#=x(x-4)#

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